Show that in every sequence of 79 consecutive positive integers, there is a positive integer whose sum of digits is divisible by 13.
(In reply to
proof (but not a proof that this is the lowest achievable) by Charlie)
The triggering of a change in the higher-order digits' sum mod 13 is ordinarily 1, except when the last digit is 9. In that case the mod-13 value decreases by 8 or increases by 5, again, except when the next digit to the left is again 9, etc. If x represents a set of digits not ending in 9, then x9 results in +5; x99 represents +9; x999, +0; x9999, +4;x99999,+8; x999999,+12;x9999999,+3;x99999999, +7; what we are looking for. The last two digits of the first number are 60, with a mod-13 value of 6, while 99999999 has a value of 7, so the value of the last ten digits is 13 or zero and we don't need to add more. So, if I'm right 9999999960 would be the first case where the next sod with a value of 0 mod 13 would be 79 higher, at 10000000039. The sod's of each of these numbers is in fact divisible by 13. Finding the mod values of numbers between, shows
n sod q r where q is the quotient when sod is divided by 13 and r the remainder
9999999960 78 6 0
9999999961 79 6 1
9999999962 80 6 2
9999999963 81 6 3
9999999964 82 6 4
9999999965 83 6 5
9999999966 84 6 6
9999999967 85 6 7
9999999968 86 6 8
9999999969 87 6 9
9999999970 79 6 1
9999999971 80 6 2
9999999972 81 6 3
9999999973 82 6 4
9999999974 83 6 5
9999999975 84 6 6
9999999976 85 6 7
9999999977 86 6 8
9999999978 87 6 9
9999999979 88 6 10
9999999980 80 6 2
9999999981 81 6 3
9999999982 82 6 4
9999999983 83 6 5
9999999984 84 6 6
9999999985 85 6 7
9999999986 86 6 8
9999999987 87 6 9
9999999988 88 6 10
9999999989 89 6 11
9999999990 81 6 3
9999999991 82 6 4
9999999992 83 6 5
9999999993 84 6 6
9999999994 85 6 7
9999999995 86 6 8
9999999996 87 6 9
9999999997 88 6 10
9999999998 89 6 11
9999999999 90 6 12
10000000000 1 0 1
10000000001 2 0 2
10000000002 3 0 3
10000000003 4 0 4
10000000004 5 0 5
10000000005 6 0 6
10000000006 7 0 7
10000000007 8 0 8
10000000008 9 0 9
10000000009 10 0 10
10000000010 2 0 2
10000000011 3 0 3
10000000012 4 0 4
10000000013 5 0 5
10000000014 6 0 6
10000000015 7 0 7
10000000016 8 0 8
10000000017 9 0 9
10000000018 10 0 10
10000000019 11 0 11
10000000020 3 0 3
10000000021 4 0 4
10000000022 5 0 5
10000000023 6 0 6
10000000024 7 0 7
10000000025 8 0 8
10000000026 9 0 9
10000000027 10 0 10
10000000028 11 0 11
10000000029 12 0 12
10000000030 4 0 4
10000000031 5 0 5
10000000032 6 0 6
10000000033 7 0 7
10000000034 8 0 8
10000000035 9 0 9
10000000036 10 0 10
10000000037 11 0 11
10000000038 12 0 12
10000000039 13 1 0
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Posted by Charlie
on 2017-08-12 12:36:27 |
Existence proof (lowest pair requiring 79)
